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Possible Duplicate:
Cartesian product

I'm Haskell newbie and I have a problem. I want to do some function that will take first element of list and connect to all elements of second list, after that take second element from first list and do the same. For example I want to take: [[1],[2],[3]) and [[4],[5],[6]] and get in output

[([1],[4]),([1],[5]),([1],[6]),
([2],[4]),([2],[5]),([2],[6]),
([3],[4]),([3],[5]),([3],[6])]

The closes one I found is transpose 

transpose [[1,2,3],[4,5,6]]
[[1,4],[2,5],[3,6]]

I would appreciate any help.

Edit: Shame on me. I found solution 

[[x,y] | x <- [[1],[2],[3]], y <- [[4],[5],[6]]]

Which result is: 

[[[1],[4]],[[1],[5]],[[1],[6]],[[2],[4]],[[2],[5]],[[2],[6]],[[3],[4]],[[3],[5]],[[3],[6]]]
Rodrigo de Azevedo
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ahaw
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3 Answers3

4
import Control.Applicative

(,) <$> [[1],[2],[3]] <*> [[4],[5],[6]]

--[([1],[4]),([1],[5]),([1],[6]),([2],[4]),([2],[5]),([2],[6]),([3],[4]),([3],[5]),([3],[6])]

See http://learnyouahaskell.com/functors-applicative-functors-and-monoids#applicative-functors for an explanation.

You can also use do-Notation, as lists are not only Applicative, but Monads, too:

do x<-[[1],[2],[3]]; y<-[[4],[5],[6]]; return (x,y)

--[([1],[4]),([1],[5]),([1],[6]),([2],[4]),([2],[5]),([2],[6]),([3],[4]),([3],[5]),([3],[6])]
Landei
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1

I'm also new to haskell, here is my solution to your question, hope it's helpful:

f [] _ = []
f (x:xs) ys = zip (take (length ys) (repeat x)) ys ++ f xs ys

I think the code explains itself quite straight forward :)

bearzk
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1

This is interesting.

sequence [[[1],[2],[3]] , [[4],[5],[6]]]
Lost dog
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